(a) Field of the Invention
The present invention relates to a coding method in a communication system. More specifically, the present invention relates to a method for generating a parity-check matrix of low-density parity-check (LDPC) codes and coding the low-density parity-check LDPC codes.
(b) Description of the Related Art
Much research and development has been carried out on high speed and high data rate transmission methods as requirements of a broadband multimedia wireless service in a wireless communication system have significantly increased. An LDPC coding method is a method for increasing data rates by efficiently correcting data transmission errors caused by distortion of signals, and the method has been researched and developed in relation to a 4th generation mobile communication system, a satellite communication system, and a high-speed wireless local area network (LAN).
The LDPC coding method is one of block coding methods using matrixes before transmitting information bits. A block coding operation is performed by using a coding matrix having few non-zero elements in the LDPC coding method. A parity-check matrix H is used as the coding matrix, and the parity-check matrix H is a large size sparse matrix having few non-zero elements.
The parity-check matrix H has a relation with code data C as shown in Equation 1.H*CT=0  [Equation 1]
where code data C includes information data S and parity data P, that is, C=[S P].
Various LDPC coding methods have been suggested to calculate the parity data P.
The parity-check matrix H is divided into sub-matrixes [A B] (that is, H=[A B]), and the parity data P is calculated by using Equation 1 as shown in Equation 2 without calculating a code generator matrix in a conventional coding method.H*CT=[A B]*[S P]T=0AST=BPT PT=B−1AST  [Equation 2]
At this time, an inverse matrix operation of the sub-matrix B of the parity-check matrix H is required. Therefore, the LDPC coding operation becomes complicated and coding speed is reduced.
In another conventional LDPC coding method, the parity-check matrix H is formed as shown in FIG. 1 in order to efficiently perform the LDPC coding operation, and a sub-matrix operation of the parity-check matrix H is simplified to calculate the parity data.
                                          H            en                    =                      [                                                            A                                                  B                                                  T                                                                              C                                                  D                                                  E                                                      ]                          ⁢                                  ⁢        H        =                                            [                                                                    I                                                        0                                                                                                              -                                              ET                                                  -                          1                                                                                                                          I                                                              ]                        ⋆                          H              en                                =                                                 [                                                                    A                                                        B                                                        T                                                                                                                                                                -                                                      ET                                                          -                              1                                                                                                      ⁢                        A                                            +                      C                                                                                                                                                    -                                                      ET                                                          -                              1                                                                                                      ⁢                        B                                            +                      D                                                                            0                                                              ]                                                          [                  Equation          ⁢                                          ⁢          3                ]            
where a sub-matrix T is a lower triangular matrix.
When the code data C is [S P1 P2], the parity data is calculated by using Equations 1 and 3 as shown in Equation 4.
                              H          ⋆                      C            T                          =                                                                                         [                                                                                    A                                                                    B                                                                    T                                                                                                                                                                                                -                                                              ET                                                                  -                                  1                                                                                                                      ⁢                            A                                                    +                          C                                                                                                                                                                                -                                                              ET                                                                  -                                  1                                                                                                                      ⁢                            B                                                    +                          D                                                                                            0                                                                              ]                                ⋆                                                      [                                          S                      ⁢                                                                                          ⁢                                              P                                                  1                          ⁢                                                                                                                                                    ⁢                                              P                        2                                                              ]                                    T                                            =                                                0                  ⁢                                                                          ⁢                                                            AS                      T                                        +                                          BP                      1                      T                                        +                                          TP                      2                      T                                                                      =                                                      0                    ⁢                                                                                  ⁢                                                                                            (                                                                                                                    -                                                                  ET                                                                      -                                    1                                                                                                                              ⁢                              A                                                        +                            C                                                    )                                                ⁢                        ST                                            +                                                                        (                                                                                                                    -                                                                  ET                                                                      -                                    1                                                                                                                              ⁢                              B                                                        +                            D                                                    )                                                ⁢                                                  P                          1                          T                                                                                                      =                                      0                    ⁢                                                                                  ⁢                    where                                                                        ,                          ϕ              =                                                                                          -                                              ET                                                  -                          1                                                                                      ⁢                    B                                    +                                      D                    ⁢                                                                                  ⁢                                          P                      1                      T                                                                      =                                                                            -                                                                        ϕ                                                      -                            1                                                                          ⁡                                                  (                                                                                                                    -                                                                  ET                                                                      -                                    1                                                                                                                              ⁢                              A                                                        +                            C                                                    )                                                                                      ⁢                    ST                    ⁢                                                                                  ⁢                                          P                      2                      T                                                        =                                                                                    -                                                                              T                                                          -                              1                                                                                ⁡                                                      (                                                                                          AS                                T                                                            +                                                              BP                                1                                T                                                                                      )                                                                                              ⁢                                                                                          ⁢                      C                                        =                                          [                                              S                        ⁢                                                                                                  ⁢                                                  P                          1                                                ⁢                                                                                                  ⁢                                                  P                          2                                                                    ]                                                                                                                              [                  Equation          ⁢                                          ⁢          4                ]            
In this LDPC coding method, an inverse matrix Φ−1 of a matrix Φ calculated by the sub-matrix operation of the parity-check matrix H is required to be calculated. The inverse matrix Φ−1 is not calculated when a determinant of the matrix Φ is 0. Accordingly, a problem exists in that the parity-check matrix H is required to be regenerated.
Parity-check matrixes shown in FIG. 2A and FIG. 2B may be used in another conventional method for performing the parity LDPC coding operation without calculating the inverse matrix of the sub-matrix of the parity-check matrix H.
As shown in FIG. 2A and FIG. 2B, the parity-check matrix H includes a [p×p] commutative matrix a. I is a [p×p] unit matrix, O is a [p×p] zero matrix, and a is a matrix moved by k from the [p×p] unit matrix. For example, when p is 5, a is given as below.
  a  =      [                            0                          1                          0                          0                          0                                      0                          0                          1                          0                          0                                      0                          0                          0                          1                          0                                      0                          0                          0                          0                          1                                      1                          0                          0                          0                          0                      ]  
At this time, the parity-check matrix H is divided into the sub-matrixes [A B] (that is, H=[A B]), and the sub-matrix A is a triangular matrix. Accordingly, a generator matrix is not required to be generated, and no inverse matrix calculation is required. In this case, the parity data is calculated by an operation according to Equation 5.H*CT=[A B]*[P S]T=0APT=BST=X  [Equation 5]
The parity data P is calculated by a back substitution by using the triangular matrix A. However, a problem exists in that a memory is required to store the sub-matrix A, and an additional operation for the parity calculation by the back substitution is also required.